Worked example 1
Step 1 (fast, reversible): 2 NO ⇌ N₂O₂. Step 2 (slow): N₂O₂ + O₂ → 2 NO₂. Derive the overall rate law.
Try it first: Write the slow step rate law first and note which species in it is an intermediate.
0 of 4 steps revealed.
Chemical KineticsReaction MechanismsContent level: Challenge 28 min
Fast Pre-Equilibria, Intermediates and Catalysis
What you'll be able to do: Derive a rate law when the rate-determining step follows a fast equilibrium, and explain how catalysts alter mechanisms.
Best after: Elementary Steps and the Rate-Determining Step
Introduction
When the slow step comes second, its rate law contains an intermediate, and intermediates cannot appear in a final rate law because their concentrations are not experimentally controlled. The fast pre-equilibrium approximation is the standard way out.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Apply the fast pre-equilibrium approximation to eliminate an intermediate
- Derive a multistep rate law and compare it with experimental data
- Explain how homogeneous and heterogeneous catalysts alter a mechanism
- Interpret negative or fractional orders in mechanistic terms
Lesson
Why intermediates must be eliminated
A valid rate law contains only species whose concentrations an experimenter can set. Intermediates exist at tiny, self-adjusting concentrations, so any rate law containing one is incomplete and must be rewritten.
The pre-equilibrium approximation
If step 1 is fast and reversible and step 2 is slow, step 1 stays effectively at equilibrium throughout. Setting the forward rate equal to the reverse rate gives an expression for the intermediate in terms of ordinary reactants, which is then substituted into the slow step.
k₁[A][B] = k-1[I] => [I] = (k₁/k-1)[A][B]
A worked pattern
With step 1 fast: 2 NO ⇌ N₂O₂ and step 2 slow: N₂O₂ + O₂ → 2 NO₂, the slow step gives rate = k₂[N₂O₂]. Equilibrium in step 1 gives [N₂O₂] = (k₁/k-1)[NO]². Substituting produces rate = k[NO]²[O₂], with k a combination of the three individual constants.
rate = (k₂ k₁ / k-1) [NO]² [O₂]
Fractional and negative orders
A negative order arises when a product of the fast step appears in the denominator of the intermediate expression, so adding that product suppresses the rate. Fractional orders often signal an equilibrium involving dissociation, such as a diatomic splitting into atoms. Neither result is an error; both are mechanistic evidence.
Catalysis inside a mechanism
A homogeneous catalyst is in the same phase as the reactants and appears explicitly in the steps, consumed early and regenerated later, which is why it can appear in the rate law. A heterogeneous catalyst works on a surface through adsorption, reaction and desorption, and saturation of that surface is the usual cause of observed zero-order behaviour.
Enzymes are biological catalysts that show exactly this pattern: first order in substrate when dilute, zero order once saturated.
Error analysis
Two mistakes dominate. The first is leaving an intermediate in the final rate law, which makes the answer untestable. The second is assuming the pre-equilibrium approximation applies when the first step is not fast and reversible; in that case a steady-state treatment is needed instead, which is beyond the scope of this course but worth knowing exists.
Key ideas
Rule
No intermediates in the answer
A final rate law may contain only reactants, products or catalysts, never intermediates.
Rule
Pre-equilibrium condition
Set forward rate equal to reverse rate for the fast step and solve for the intermediate.
Definition
Homogeneous catalyst
A catalyst in the same phase as the reactants; it can appear in the rate law.
Definition
Heterogeneous catalyst
A catalyst in a different phase, acting by adsorption on a surface.
Key concept
Negative order
A species that suppresses the rate, typically a product of a fast reversible step.
Equation
Pre-equilibrium substitution
[I] = (k₁/k-1) × (reactant terms of the fast step)
- k₁ = forward rate constant of the fast step
- k-1 = reverse rate constant of the fast step
Equation
Observed rate constant
k(obs) = k₂ k₁ / k-1
- k₂ = rate constant of the slow step
Worked examples
Worked example 2
A mechanism has step 1 fast and reversible: A ⇌ B + C, and step 2 slow: B + D → E. Show why the rate is inversely proportional to [C].
0 of 3 steps revealed.
Worked example 3
A surface-catalysed decomposition is first order in reactant at low pressure and zero order at high pressure. Explain.
0 of 3 steps revealed.
Common mistakes
Leaving an intermediate in the final rate law.
Why it's wrong: Its concentration cannot be set or measured, so the law cannot be tested.
Check instead: Eliminate it using the fast equilibrium before quoting an answer.
Assuming a catalyst can never appear in a rate law.
Why it's wrong: A homogeneous catalyst is consumed in an early step, so its concentration can control the rate.
Check instead: Catalysts may appear in the rate law even though they are not consumed overall.
Treating a fractional or negative order as a mistake in the data.
Why it's wrong: Both are natural consequences of pre-equilibria.
Check instead: Look for a fast reversible step that explains the unusual order.
Applying the pre-equilibrium approximation when the first step is slow.
Why it's wrong: The approximation depends on the first step being fast and reversible.
Check instead: Check the step labels before choosing the method.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
If a fast reversible step precedes the rate-determining step, the fast step is treated as being at equilibrium, so the forward and reverse rates are equal. Solving that equality for the intermediate concentration and substituting it into the slow step's rate law eliminates the intermediate and produces a rate law in measurable species only. This procedure explains fractional and negative orders and, together with catalysis, accounts for most rate laws that look impossible at first glance. Catalysts open a lower-energy pathway and are regenerated, so they appear in the mechanism and can appear in the rate law without being consumed overall.
- Final rate laws must not contain intermediates
- Set forward equal to reverse for a fast step and solve for the intermediate
- Substituting gives an observed k that combines several rate constants
- Negative and fractional orders are mechanistic evidence, not errors
- Catalysts appear in mechanisms and may appear in rate laws without being consumed
Sources and further reading
- Chemistry 2e, Section 12.6: Reaction MechanismsPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 12.6View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License
- Chemistry 2e, Section 12.7: CatalysisPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 12.7View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License